In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $$CH_3OH(\ell) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(\ell)$$ At $$298$$ K standard Gibb's energies of formation for $$CH_3OH(\ell), H_2O(\ell)$$ and $$CO_2(g)$$ are $$-166.2, -237.2$$ and $$-394.4$$ kJ mol$$^{-1}$$ respectively. If standard enthalpy of combustion of methanol is $$-726$$ kJ mol$$^{-1}$$, efficiency of the fuel cell will be

The efficiency of a fuel cell is defined as the ratio of the maximum useful electrical work obtainable (equal to the decrease in Gibbs free energy, $$-\Delta G^\circ$$) to the heat released at constant pressure, i.e. the enthalpy of combustion $$-\Delta H^\circ$$:

$$\text{Efficiency } (\eta)=\frac{-\Delta G^\circ}{-\Delta H^\circ}\times 100\%$$

Step 1 : Calculate the standard Gibbs free energy change, $$\Delta G^\circ$$

Reaction: $$CH_3OH(\ell)+\frac32 O_2(g)\rightarrow CO_2(g)+2H_2O(\ell)$$

Standard Gibbs energies of formation (in kJ mol$$^{-1}$$):
$$\Delta_fG^\circ(CH_3OH)=-166.2,\quad \Delta_fG^\circ(H_2O)=-237.2,\quad \Delta_fG^\circ(CO_2)=-394.4,\quad \Delta_fG^\circ(O_2)=0$$

Using $$\Delta G^\circ =\sum \Delta_fG^\circ(\text{products})-\sum \Delta_fG^\circ(\text{reactants})$$:

$$\Delta G^\circ =\bigl[-394.4 + 2(-237.2)\bigr] -\bigl[-166.2 + \tfrac32(0)\bigr]$$

$$\Delta G^\circ =\bigl[-394.4-474.4\bigr] -(-166.2)=(-868.8)+166.2$$

$$\Delta G^\circ = -702.6\text{ kJ mol}^{-1}$$

Step 2 : Use the given enthalpy of combustion, $$\Delta H^\circ$$

The standard enthalpy of combustion of methanol is provided as
$$\Delta H^\circ = -726\text{ kJ mol}^{-1}$$

Step 3 : Compute the efficiency

$$\eta=\frac{-\Delta G^\circ}{-\Delta H^\circ}\times 100\% =\frac{702.6}{726}\times 100\%$$

$$\eta = 0.9684 \times 100\% \approx 96.8\% \approx 97\%$$

Therefore, the efficiency of the methanol-oxygen fuel cell is about $$97\%$$.

Option D which is: $$97\%$$